Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $r \neq 0$. $n = \dfrac{r + 7}{r + 5} \times \dfrac{-8r - 40}{r^2 + 16r + 63} $
Solution: First factor the quadratic. $n = \dfrac{r + 7}{r + 5} \times \dfrac{-8r - 40}{(r + 7)(r + 9)} $ Then factor out any other terms. $n = \dfrac{r + 7}{r + 5} \times \dfrac{-8(r + 5)}{(r + 7)(r + 9)} $ Then multiply the two numerators and multiply the two denominators. $n = \dfrac{ (r + 7) \times -8(r + 5) } { (r + 5) \times (r + 7)(r + 9) } $ $n = \dfrac{ -8(r + 7)(r + 5)}{ (r + 5)(r + 7)(r + 9)} $ Notice that $(r + 5)$ and $(r + 7)$ appear in both the numerator and denominator so we can cancel them. $n = \dfrac{ -8\cancel{(r + 7)}(r + 5)}{ (r + 5)\cancel{(r + 7)}(r + 9)} $ We are dividing by $r + 7$ , so $r + 7 \neq 0$ Therefore, $r \neq -7$ $n = \dfrac{ -8\cancel{(r + 7)}\cancel{(r + 5)}}{ \cancel{(r + 5)}\cancel{(r + 7)}(r + 9)} $ We are dividing by $r + 5$ , so $r + 5 \neq 0$ Therefore, $r \neq -5$ $n = \dfrac{-8}{r + 9} ; \space r \neq -7 ; \space r \neq -5 $